Question: $\begin{aligned} &C = 3y^2+4y+4 \\\\ &D = -7y^2+3y-6 \end{aligned}$ $D-C=$
Explanation: Since we are asked to find $D-C$, let's substitute in the trinomial expressions that we are given for $D$ and $C$ : $D-C = (-7y^2+3y-6)-(3y^2+4y+4)$ Since we are subtracting, it is helpful to distribute the $\text{{negative sign}}$ across all terms in the second trinomial: $\begin{aligned}&(-7y^2+3y-6){-}(3y^2+4y+4)\\ \\ =&(-7y^2+3y-6){-}3y^2{-}4y{-}4\\ \\ =&-7y^2+3y-6-3y^2-4y-4 \end{aligned}$ Note that the parentheses around the first trinomial don't affect the order of operations, so we can just remove them. When we add or subtract terms in a polynomial expression, the only way that we can simplify the expression is by combining those terms that are alike. Our expression contains terms of $3$ different degrees in the same variable: ${y^2}, {y},$ and the $\text{{constant}}$ term: ${{-7y^2} {+3y} {-6} {-3y^2} {-4y} {-4}}$ Now that we have identified like terms, let's combine them. Make sure to keep track of positive and negative signs! ${{(-7-3)y^2} + {(3-4)y} + {(-6-4)}}$ When we add the coefficients in front of each term, we get the following trinomial: ${-10y^2-y-10}$